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What condition on \({\rm{\alpha and \beta }}\) is necessary for the standard beta pdf to be symmetric?

Short Answer

Expert verified

\({\rm{\alpha = \beta }}\)

Step by step solution

01

Definition of probability

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the \({\rm{\alpha  and \beta }}\) is necessary for the standard beta pdf to be symmetric

A typical beta distribution with parameters \({\rm{\alpha and \beta }}\) is assumed for a random variable \({\rm{X}}\). We know that \({\rm{X}}\) pdf will be non-zero only in the range 0£x£1.

As a result, the pdf's mean should be \({\rm{0}}{\rm{.5}}\) in order for it to be symmetric. We also know that the standard beta distribution \({\rm{\mu }}\)is as follows:

\(\begin{aligned}{}{\rm{\mu }}&{{\rm{ = }}\frac{{\rm{\alpha }}}{{\rm{\beta }}}}\\{\frac{{\rm{1}}}{{\rm{2}}}}&{{\rm{ = }}\frac{{\rm{\alpha }}}{{{\rm{\alpha + \beta }}}}}\\{{\rm{\alpha + \beta }}}&{{\rm{ = 2\alpha }}}\\{\rm{\alpha }}&{{\rm{ = \beta }}}\end{aligned}\)

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Most popular questions from this chapter

The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf

\(f(x)= \left\{ {\begin{array}{*{20}{c}}{\frac{3}{2} \times \frac{1}{{{x^2}}}}&{1£x£3} \\0&{{\text{ }}otherwise{\text{ }}}\end{array}} \right.\)

a. Obtain the cdf.

b. What is the probability that reaction time is at most\({\rm{2}}{\rm{.5sec}}\)? Between \({\rm{1}}{\rm{.5}}\) and\({\rm{2}}{\rm{.5sec}}\)?

c. Compute the expected reaction time.

d. Compute the standard deviation of reaction time.

e. If an individual takes more than \({\rm{1}}{\rm{.5sec}}\) to react, a light comes on and stays on either until one further second has elapsed or until the person reacts (whichever happens first). Determine the expected amount of time that the light remains lit. (Hint: Let \({\rm{h(X) = }}\) the time that the light is on as a function of reaction time\({\rm{X}}\).)

In each case, determine the value of the constant\({\rm{c}}\)that makes the probability statement correct. a.\({\rm{\Phi (c) = }}{\rm{.9838}}\)b.\({\rm{P(0}} \le {\rm{Z}} \le {\rm{c) = }}{\rm{.291}}\)c.\({\rm{P(c}} \le {\rm{Z) = }}{\rm{.121}}\)d.\({\rm{P( - c}} \le {\rm{Z}} \le {\rm{c) = }}{\rm{.668}}\)e.\({\rm{P(c}} \le {\rm{|Z|) = }}{\rm{.016}}\)

The accompanying sample consisting of observations \({\rm{n = 20}}\) on dielectric breakdown voltage of a piece of epoxy resin appeared in the article "Maximum Likelihood Estimation in the \({\rm{3}}\)-Parameter Weibull Distribution (IEEE Trans. on Dielectrics and Elec. Insul., \({\rm{1996}}\): \({\rm{43 - 55)}}\). The values of \({\rm{(i - }}{\rm{.5)/n}}\) for which \({\rm{z}}\) percentiles are needed are \({\rm{(1 - }}{\rm{.5)/20 = }}{\rm{.025,(2 - }}{\rm{.5)/20 = }}\)\({\rm{.075, \ldots }}\), and\({\rm{.975}}\). Would you feel comfortable estimating population mean voltage using a method that assumed a normal population distribution?

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{24}}{\rm{.46}}}&{{\rm{25}}{\rm{.61}}}&{{\rm{26}}{\rm{.25}}}&{{\rm{26}}{\rm{.42}}}&{{\rm{26}}{\rm{.66}}}\\{{\rm{ zpercentile }}}&{{\rm{ - 1}}{\rm{.96}}}&{{\rm{ - 1}}{\rm{.44}}}&{{\rm{ - 1}}{\rm{.15}}}&{{\rm{ - }}{\rm{.93}}}&{{\rm{ - }}{\rm{.76}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.15}}}&{{\rm{27}}{\rm{.31}}}&{{\rm{27}}{\rm{.54}}}&{{\rm{27}}{\rm{.74}}}&{{\rm{27}}{\rm{.94}}}\\{{\rm{ zpercentile }}}&{{\rm{ - }}{\rm{.60}}}&{{\rm{ - }}{\rm{.45}}}&{{\rm{ - }}{\rm{.32}}}&{{\rm{ - }}{\rm{.19}}}&{{\rm{ - }}{\rm{.06}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.98}}}&{{\rm{28}}{\rm{.04}}}&{{\rm{28}}{\rm{.28}}}&{{\rm{28}}{\rm{.49}}}&{{\rm{28}}{\rm{.50}}}\\{{\rm{ zpercentile }}}&{{\rm{.06}}}&{{\rm{.19}}}&{{\rm{.32}}}&{{\rm{.45}}}&{{\rm{.60}}}\\{{\rm{ Observation }}}&{{\rm{28}}{\rm{.87}}}&{{\rm{29}}{\rm{.11}}}&{{\rm{29}}{\rm{.13}}}&{{\rm{29}}{\rm{.50}}}&{{\rm{30}}{\rm{.88}}}\\{{\rm{ zpercentile }}}&{{\rm{.76}}}&{{\rm{.93}}}&{{\rm{1}}{\rm{.15}}}&{{\rm{1}}{\rm{.44}}}&{{\rm{1}}{\rm{.96}}}\end{array}\)

Let Z have a standard normal distribution and define a new rv Y by \[{\text{Y = \sigma Z + \mu }}\]. Show that Y has a normal distribution with parameters \[{\text{\mu }}\] and \[{\text{\sigma }}\]. (Hint: \[{\text{Y\poundsy}}\]if \[{\text{Z\pounds}}\]? Use this to find the cdf of Y and then differentiate it with respect to y.)

Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with a mean of\({\bf{70}}\)and a standard deviation of\({\bf{3}}\). a. If a specimen is acceptable only if its hardness is between 67 and 75, what is the probability that a randomly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is\(\left( {{\bf{70}} - {\bf{c}},{\rm{ }}{\bf{70}} + {\bf{c}}} \right)\), for what value of c would\({\bf{95}}\% \)of all specimens have acceptable hardness? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than\({\bf{73}}.{\bf{84}}\)?

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